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Chapter 16 - 16.96

96. The distance from the relaxed position of the bottom end of the spring to its equilibrium position when

the body is attached is given by Hooke’s law: ∆ x = F/k =(0 . 20kg)(9 . 8m / s 2 ) / (19N / m) = 0 . 103 m.

(a) The body, once released, will not only fall through the ∆ x distance but continue through the

equilibrium position to a “turning point” equally far on the other side. Thus, the total descent of

the body is 2∆ x =0 . 21 m.

(b) Since f = ω/ 2 π , Eq. 16-12 leads to

f = 1

2 π

k

m =1 . 6Hz .

(c) The maximum distance from the equilibrium position is the amplitude: x m =∆ x =0 . 10 m.